Wires A and B are made from the same material | vedclass

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(B) The energy stored in a stretched wire is given by $W = \frac{1}{2} \frac{F^2 L}{AY}$,where $F$ is the tension,$L$ is the length,$A$ is the cross-s

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$3 : 4$

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(B) The energy stored in a stretched wire is given by $W = \frac{1}{2} \frac{F^2 L}{AY}$,where $F$ is the tension,$L$ is the length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.Since the wires are made of the same material,$Y_A = Y_B = Y$. The tension $F$ is the same for both.Thus,$W \propto \frac{L}{A}$.Since $A = \pi r^2$,we have $W \propto \frac{L}{r^2}$.Given $L_A = 3L_B$ and $d_A = 2d_B \implies r_A = 2r_B$.Therefore,$\frac{W_A}{W_B} = \frac{L_A}{L_B} 	imes \left(\frac{r_B}{r_A}ight)^2$.Substituting the values: $\frac{W_A}{W_B} = 3 	imes \left(\frac{1}{2}ight)^2 = 3 	imes \frac{1}{4} = \frac{3}{4}$.

Wires $A$ and $B$ are made from the same material. $A$ has twice the diameter and three times the length of $B$. If the elastic limits are not reached,when each is stretched by the same tension,the ratio of energy stored in $A$ to that in $B$ is

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